A hyperbola,having the transverse axis of len | vedclass

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(B) Given,the length of the transverse axis of the hyperbola is $2 a_{1} = 2 \sin 	heta$,so $a_{1} = \sin 	heta$.For the ellipse $3 x^{2} + 4 y^{2}

Vedclass · Accepted Answer

$x^{2} \operatorname{cosec}^{2} 	heta-y^{2} \sec ^{2} 	heta=1$

Vedclass · Answer

(B) Given,the length of the transverse axis of the hyperbola is $2 a_{1} = 2 \sin 	heta$,so $a_{1} = \sin 	heta$.For the ellipse $3 x^{2} + 4 y^{2} = 12$,we divide by $12$ to get $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$.Here,$a^{2} = 4$ and $b^{2} = 3$.The eccentricity $e$ of the ellipse is given by $b^{2} = a^{2}(1 - e^{2})$,so $3 = 4(1 - e^{2})$,which gives $e^{2} = 1 - \frac{3}{4} = \frac{1}{4}$,so $e = \frac{1}{2}$.The focus of the ellipse is $(\pm ae, 0) = (\pm 2 	imes \frac{1}{2}, 0) = (\pm 1, 0)$.Since the hyperbola is confocal with the ellipse,its focus is $(\pm 1, 0)$.For the hyperbola,$a_{1} e_{1} = 1$. Substituting $a_{1} = \sin 	heta$,we get $e_{1} = \frac{1}{\sin 	heta} = \operatorname{cosec} 	heta$.Now,$b_{1}^{2} = a_{1}^{2}(e_{1}^{2} - 1) = a_{1}^{2} e_{1}^{2} - a_{1}^{2} = 1 - \sin^{2} 	heta = \cos^{2} 	heta$.The equation of the hyperbola is $\frac{x^{2}}{a_{1}^{2}} - \frac{y^{2}}{b_{1}^{2}} = 1$,which becomes $\frac{x^{2}}{\sin^{2} 	heta} - \frac{y^{2}}{\cos^{2} 	heta} = 1$.This simplifies to $x^{2} \operatorname{cosec}^{2} 	heta - y^{2} \sec^{2} 	heta = 1$.

$A$ hyperbola,having the transverse axis of length $2 \sin \theta$,is confocal with the ellipse $3 x^{2}+4 y^{2}=12$. Its equation is

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